Answered

When a 2.88 g sample of solid ammonium nitrate, NH4NO3, dissolved in 50.0 g of water on a coffee cup calorimeter, the temperature decreases from 23.0 °C to 18.4 °C. Calculate the molar enthalpy of solvation, ΔsolvHm. The molar mass of ammonium nitrate is 80.06 g/mol.



Answer :

subhamjrp

Explanation:

First, calculate the mass of water:

m_water = 50.0 g

Next, calculate the temperature change (ΔT):

ΔT = T_final - T_initial = 18.4 °C - 23.0 °C = -4.6 °C

Since the temperature decreases, the process is endothermic (heat is absorbed). The heat absorbed (q) is:

q = mcΔT

where m is the mass of water, c is the specific heat capacity of water (approximately 4.184 J/g°C), and ΔT is the temperature change:

q = 50.0 g × 4.184 J/g°C × (-4.6 °C) = -966.4 J

Since the process is endothermic, the enthalpy change (ΔH) is positive:

ΔH = q = 966.4 J

Next, calculate the number of moles of ammonium nitrate (n):

n = m / M

where m is the mass of ammonium nitrate (2.88 g) and M is the molar mass (80.06 g/mol):

n = 2.88 g / 80.06 g/mol = 0.036 mol

Finally, calculate the molar enthalpy of solvation (ΔsolvHm):

ΔsolvHm = ΔH / n = 966.4 J / 0.036 mol = 26,822 J/mol = 26.82 kJ/mol

Therefore, the molar enthalpy of solvation (ΔsolvHm) is 26.82 kJ/mol.

Other Questions